Given,
\(x^{'}=\cfrac{x-vt}{\sqrt{1-v^2}}\)
what happens when \(v\ge1\)?
When \(x^{'}\) was set to \(1\) in the triangle (post "Thought Experiment Gone Wrong" dated 20 Nov 2017),
\(cos(\theta)=\cfrac{\sqrt{1-v^2}}{1}\)
we are attempting to transform \(x^{'}\) to \(x\) where it is calibrated to unit length. We are saying if \(x^{'}\) is to be one unit what do we need to do to \(x\). Using the triangle restricts the value of \(v\) to less than \(1\).
So, given \(v\) and \(x\) and \(v\ge 1\) to use the transform, both \(v\) and \(x\) are scale up to the value of the highest speed possible, light speed \(c\) and \(|c|\) respectively;
\(\cfrac{v}{c}\rightarrow v\)
\(\cfrac{x}{|c|}\rightarrow x\)
and \(x^{'}\) is also measured in the unit of \(c\) per unit time.
\(\cfrac{x^{'}}{|c|}\rightarrow x^{'}\)
And hence Lorentz's transform.
Note: If \(x=10\,m\) then to used Lorentz transform we need \(\cfrac{10}{299792458}\rightarrow x\) instead of just \(x=10\).
